answersLogoWhite

0

0.45 repeated = 5/11

METHOD #1

0.45 repeated = 0.45 + 0.0045 + 0.000045 ...

r = 0.01 [common ratio between each term]

n = 0.45 [beginning term]

As |r|<1

0.45 + 0.0045 + 0.000045 ... forms a limiting sum

Limiting sum = S(n) = n / (1 - r)

= 0.45/(1 - 0.01)

= 0.45/0.99

= 45/99

= 5/11

METHOD #2

Let x = 0.45 repeated

100x = 45.45 repeated

100x - x = 45.454545454545... - 0.454545454545...

99x = 45

x = 45/99

x = 5/11

User Avatar

Wiki User

13y ago

What else can I help you with?